Question

The value of
\( \dfrac{3}{4} + \dfrac{15}{16} + \dfrac{63}{64} + \cdots \)
up to \( n \) terms is

• A. \( n-\dfrac{4^n-1}{3} \)
• B. \( n+\dfrac{4^n-1}{3} \)
• C. \( n+\dfrac{4^n}{3}-\dfrac13 \)
• D. n-(4ⁿ)/(3)+\dfrac13

Hint:

Rewrite complex series as difference of simpler sums.

Answer 1

The Correct Option is C

The series can be written as: sumk=1ⁿ (1-(1)/(4ᵏ)) = n-(1)/(3)+(4ⁿ)/(3)